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4

Essentially, the problem is $u''+\lambda u=0$ on $[-\pi,\pi]$ with periodic boundary conditions. (Note that this interval has the same length as the circumference of the circle; this choice of parametrization ensures that the Laplace-Beltrami operator on the circle directly corresponds to the Laplacian on this interval.) The solutions to the DE itself are ...


3

You have found one particular solution, and since the PDE is linear any other solution will differ from that one by a solution of the homogeneous equation: $$ x \frac{\partial z}{\partial x}+t \frac{\partial z}{\partial t}+y \frac{\partial z}{\partial y}=0 . $$ This equation says that the directional derivative of $z$ in the radial direction is zero; in ...


2

For any function $\phi$ and coordinate variable $w$, we will use $\partial_w$ and $\phi_w$ as a shorthand of the differential operator $\frac{\partial}{\partial w}$ and the partial derivative $\partial_w \phi = \frac{\partial\phi}{\partial w}$. Consider following change of variables $$(x,y) = \left(\frac{\sqrt{3}u+v}{2}, ...


2

Have you checked if the proof to theorem A. uses weak lower-semicontinuity or sequential weak lower-semicontinuity? I would guess it is the latter, especially after reading @user127096 response in the link you provided. As for your questions, (i) Note that for $t\in\mathbb R$, $$ I(tu)=\frac{|t|^2}{2}\int|\nabla u|^2-\frac{|t|^q}{q}\int|u|^q\sim_{|t|\to0} ...


2

If the method of separation of variables is required, first you have to transform the inhomogeneous PDE to an homogeneous PDE. Any particular solution can be used. For example, obviously $U=-4y$ is a particular solution. So, let $U(x,y)=-4y+V(x,y)$ $U_y=-4+V_y \quad\to\quad U_{xy}=V_{xy}$ $4+U_y-U_{xy}=0 \quad\to\quad 4+(-4+V_y)+V_{xy}=0$ ...


2

Hint If we set $V := U_y$, we may rewrite the equation as an "upper triangular" system of what are effectively o.d.e.s: $$\left\{ \begin{array}{rcl} U_y &=& V \\ V_x &=& V + 4 \end{array} \right.$$ The second equation has general solution $$V(x, y) = g(y) e^x - 4 .$$


2

As noted in a comment, the boundary condition $f(1,t)=0$ can be accommodated using a negative image charge at $x=2$, leading to the solution $p(x,t)-p(x-2,t)$ (with $p$ the fundemantal solution you gave). To get $f'(1,t)=0$, you need a positive image charge at $x=2$, leading to the solution $f(x,t)=p(x,t)+p(x-2,t)$, with \begin{align} ...


1

The underlying problem is that the formula $\int \frac1x \,dx = \ln |x| +C$ is not quite correct. The correct general form of the antiderivatives of $1/x$ is $$\begin{cases} \ln x + C_1,\quad & x>0 \\ \ln (-x) + C_2,\quad & x<0\end{cases}$$ There's no reason for $C_1$ and $C_2$ to be the same constant. I'll explain how this is relevant, ...


1

The answer to the question raised by "sequence" is given in the comments. So, my answer is only a different form (but equivalent) of the method of characteristics with advantage of clearness : $$u_t+cu_x=e^{2x}$$ The characteristic equations are : $\quad \frac{dt}{1}=\frac{dx}{c}=\frac{du}{e^{2x}}$ From $\frac{dt}{1}=\frac{dx}{c}$ , the first ...


1

$$ x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}+t \frac{\partial z}{\partial t}=xyt$$ Solving thanks to the method of characteristics : The characteristic equations are : $$\frac{dx}{x}=\frac{dy}{y}=\frac{dt}{t}=\frac{dz}{xyt}$$ From $\frac{dx}{x}=\frac{dt}{t}$ the first characteristic : $\frac{x}{t}=c_1$ From ...


1

HINT (partial answer) : Search for a particular linear solution on the form $z=ax+by+c$ Puting into the PDE leads to $$z=ax+by+a^2+ab+b^2 \quad\text{any }a\:,\: b.$$


1

There are many ways to show that there is no unique solution to this PDE. One of the simplest we can use to argue is: "the derivative doesn't read constants" So if $u$ is a particoular integral of your PDE, then also $u+constant$ is


1

Consider $$\left\{\begin{array}{c} −\Delta u = f & \text{on }\Omega \\ u(x)=0, \partial\Omega \end{array}\right.,$$ where $\Omega \subset \mathbb{R}^{N}$ is open, bounded and $f \in L^{2}(\Omega)$ See that $u \in H^{1}_{0}(\Omega)$ is a weak solution for the problem above if $(u,v)_{H^{1}_{0}(\Omega)} = \displaystyle\int_{\Omega} \nabla u \nabla v = ...


1

For $u \in L^2$ one defines $\Delta u \in H^{-2} = (H_0^2)^*$ by $$ \langle \Delta u, v \rangle := \int u \, \Delta v \, \mathrm{d}x.$$ Using Cauchy-Schwarz, you find $$ |\langle \Delta u , v \rangle | \le \| u\|_{L^2} \, \|\Delta v\|_{L^2} \le C \, \|u\|_{L^2} \, \|v\|_{H_0^2}.$$ Hence, $\|\Delta u\|_{H^{-2}} \le C \, \|u\|_{L^2}$. This should also answer ...


1

The footnote 31 on page 305 explains the logic of the choice of $h$: the support of $w$ is at some distance from the top and side surface of the cylinder $ Q_+$, but it need not be separated from the bottom surface, which is $x_N=0$. This is how the chosen partition of unity works: in order to cover the domain by finitely many sets, we need these sets to ...


1

The Fourier Cosine series for $\sin(x)$ on $[0,\pi]$ is given by $$\sin(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos(nx)$$ where $a_n=\frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)\,dx$. Note that $a_0=\frac{4}{\pi}$ and $a_1=0$. For $n\ge 2$, we have $$\begin{align} a_n&=\frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)\,dx\\\\ &=\frac{1}{\pi}\int_0^\pi ...


1

You have two options, either do the product rule and then discretize, or discretize the equation as is. I'll derive them both: Product Rule $$\frac{du}{dt} = \frac{d}{dx}\left[\frac{1}{x^2+1} \frac{du}{dx}\right] $$ $$\frac{du}{dt} = \frac{d}{dx}\left[\frac{1}{x^2+1} \right]\frac{du}{dx} + \frac{1}{x^2+1}\frac{d^2u}{dx^2} $$ $$\frac{du}{dt} = ...


1

Assume that $u(x,y)$ is a solution then $$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$$ $$\Rightarrow \frac{\partial u}{\partial y}=\frac{du-\frac{\partial u}{\partial x}dx}{dy}$$ The equation becomes $$a\frac{\partial u}{\partial x}+b\frac{du-\frac{\partial u}{\partial x}dx}{dy}=c$$ $$a\frac{\partial u}{\partial ...


1

Not sure why you would want to find spatial periodicity separately for each cos factor. However, a general expression for a traveling wave is $U(x,t) = a\cos(2\pi)(kx-nt)$. The expression $U(x,t) = a\cos(2\pi(kx-nt)+g)$ represents another wave with phase difference $g$ relative to the first expression (Waves, C.A. Coulson). If you recast your first cos term ...


1

From the absence of the boundary conditions I assume it's a Cauchy problem. Note that $u(x,0)\to1$, $x\to+\infty$ and $u(x,0)\to-1$, $x\to-\infty$. And the convergence is exponential in speed. So it's straightforward to check that $v(x)=u(x,0)-\text{sign}(x-1)\in L_1(\mathbb R)$. Denote $\Gamma(x,t)=(4\pi t)^{-1/2}e^{-x^2/(4t)}$ the fundamental solution of ...


1

For the sake of an answer: This usually means that the solutions are less regular than the functions in the equations or the initial data.



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